The sum of three consecutive multiples of 7 is 357.

(a) Let the three consecutive multiples of 7 be $7 x,(7 x+7),(7 x+14)$

where $x$ is a natural number.

According to the question,

$7 x+(7 x+7)+(7 x+14)=357$

$\Rightarrow \quad 21 x+21=357$

$\Rightarrow \quad 21(x+1)=357$

$\Rightarrow \quad \frac{21(x+1)}{21}=\frac{357}{21}$ [dividing both sides by 21 ]

$\Rightarrow \quad x+1=17$

$\Rightarrow \quad x=17-1 \quad$ [transposing 1 to RHS]

$\therefore \quad x=16$

Hence, the smallest multiple of 7 is $7 \times 16, i . \theta, 112$.